User:IssaRice/Linear algebra/Linear transformation vs matrix views

Given an $m\times n$ matrix $A$ we can define a linear map $T:\mathbf {R} ^{n}\to \mathbf {R} ^{m}$ by $T(x)=Ax$.

Given a linear map $T:V\to W$, it is not immediately possible to get a corresponding matrix. We must choose some basis $v_{1},\ldots ,v_{n}$ for $V$ and a basis $w_{1},\ldots ,w_{m}$ for $W$. Then we can get a matrix by setting the $k$th column to be $Tv_{k}$ written in the basis $w_{1},\ldots ,w_{m}$.

We would hope that any property we attribute to a linear map is invariant of the matrix we use to represent it. For instance if $T:V\to W$ is called "injective" then it should be injective regardless of what matrix we use. Similarly given any matrix that is injective, any of the possible linear maps that that matrix represents should be injective.

Examples of other properties like this: injective, surjective, bijective, rank, diagonalizable

On the other hand, a property like "the sum of the columns is equal to such and such" is not invariant

I think the root of the confusion is that for these invariant properties, it is possible to define them given either the matrix or the map. So then there are two definitions floating around, and i don't see people showing them equivalent in general.

we can think of a linear map as an equivalence class of matrices. or we can think of a matrix as an equivalence class of linear maps. then we can phrase these invariance results as basically saying that these properties are well-defined. the difference seems to be that here we want to show equivalence, so we need to do it in both directions (?).

the actual proofs are pretty tedious (is my guess)

let's run with injectivity as an example.

Definition. A linear map $T:V\to W$ is injective iff $Tv=Tu$ implies $v=u$ for all $v,u\in V$.

Definition. An $m\times n$ matrix $A$ is injective iff $Ax=Ay$ implies $x=y$ for all $x,y\in \mathbf {R} ^{n}$.

We want to say that these are basically the same thing. How do we express that? Some ideas:

(1) if $T$ is injective and $\beta ,\gamma$ are any bases, then $A:=[T]_{\beta }^{\gamma }$ is injective

(2) if $A$ is injective and $\beta ,\gamma$ are any bases are any bases, then for any $T$ such that $A=[T]_{\beta }^{\gamma }$ we have that $T$ is injective

I think these can be combined into:

(3) for all $T,A,\beta ,\gamma$ such that $A=[T]_{\beta }^{\gamma }$: T injective iff A injective.

Potential proof strategy that might not be so tedious: basically imagine that A is always written in the standard basis. Then consider $[T]_{\sigma }^{\sigma '}$ where $\sigma ,\sigma '$ are the standard bases in R^n and R^m. Then $A=[I]_{\gamma }^{\sigma '}[T]_{\beta }^{\gamma }[I]_{\sigma }^{\beta }$.